Question: $\overline{AC}$ is $12$ units long $\overline{BC}$ is $5$ units long $\overline{AB}$ is $13$ units long What is $\sec(\angle ABC)?$ $A$ $C$ $B$ $12$ $5$ $13$
Solution: $\sec(\angle ABC) = \dfrac{1}{\cos(\angle ABC)}$ How can we find $\cos(\angle ABC)$ SOH CAH TOA osine = djacent over ypotenuse Adjacent $= \overline{BC} = 5$ Hypotenuse $= \overline{AB} = 13$ $\cos(\angle ABC) = \dfrac{5}{13}$ $\sec(\angle ABC) = \dfrac{1}{\cos(\angle ABC)} = \dfrac{13}{5}$